Elements of Geometry and Trigonometry Translated from the French of A.M. Legendre by David Brewster: Revised and Adapted to the Course of Mathematical Instruction in the United StatesA.S. Barnes & Company, 1849 - 359 sider |
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Side 5
... Measurement of Angles , Problems relating to the First and Third Books , BOOK IV . The Proportions of Figures and the Measurement of Areas , - Problems relating to the Fourth Book , 41 57 68 8888 98 BOOK V. Regular Polygons and the ...
... Measurement of Angles , Problems relating to the First and Third Books , BOOK IV . The Proportions of Figures and the Measurement of Areas , - Problems relating to the Fourth Book , 41 57 68 8888 98 BOOK V. Regular Polygons and the ...
Side 9
... measurement of extension . Extension has three dimensions , length , breadth , and height , or thickness . 2. A line is length without breadth , or thickness . The extremities of a line are called points : a point , there- fore , has ...
... measurement of extension . Extension has three dimensions , length , breadth , and height , or thickness . 2. A line is length without breadth , or thickness . The extremities of a line are called points : a point , there- fore , has ...
Side 34
... measuring unit an exact number of times , there may perhaps be a smaller unit which will be contained an exact number of times in each of the magnitudes . But if there is no unit of an assignable value , which shall be cortained an ...
... measuring unit an exact number of times , there may perhaps be a smaller unit which will be contained an exact number of times in each of the magnitudes . But if there is no unit of an assignable value , which shall be cortained an ...
Side 40
... since and or therefore , MN : P : Q RST : V MXR NXS :: P × T : Q × V MxQ = NxP Rx V = SXT , we shall have MxQxRxV = NxPxSxT MXRxQxV = NxSxP × T MXR NxS :: PXT : Q × V. BOOK III . THE CIRCLE , AND THE MEASUREMENT OF 40 GEOMETRY .
... since and or therefore , MN : P : Q RST : V MXR NXS :: P × T : Q × V MxQ = NxP Rx V = SXT , we shall have MxQxRxV = NxPxSxT MXRxQxV = NxSxP × T MXR NxS :: PXT : Q × V. BOOK III . THE CIRCLE , AND THE MEASUREMENT OF 40 GEOMETRY .
Side 41
... MEASUREMENT OF ANGLES . Definitions . 1. The circumference of a circle is a curved line , all the points of which are equally distant from a point within , called the centre . The circle is the space terminated by A this curved line ...
... MEASUREMENT OF ANGLES . Definitions . 1. The circumference of a circle is a curved line , all the points of which are equally distant from a point within , called the centre . The circle is the space terminated by A this curved line ...
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Elements of Geometry and Trigonometry Translated from the French of A.M ... Charles Davies,Adrien Marie Legendre Uten tilgangsbegrensning - 1842 |
Elements of Geometry and Trigonometry: Translated from the French of A.M ... Adrien Marie Legendre,Charles Davies Uten tilgangsbegrensning - 1848 |
Elements of Geometry and Trigonometry Translated from the French of A.M ... Charles Davies Uten tilgangsbegrensning - 1835 |
Vanlige uttrykk og setninger
adjacent altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone consequently convex surface cosine Cotang cylinder diagonal diameter dicular distance divided draw drawn equal angles equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle Scholium secant segment side BC similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex
Populære avsnitt
Side 241 - In every plane triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides is to the tangent of half their difference.
Side 251 - Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51° ; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45' ; required the height of the tower.
Side 109 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (Book IV.
Side 91 - Two similar triangles are to each other as the squares described on their homologous sides. Let ABC, DEF, be two similar triangles, having the angle A equal to D, and The angle B=E.
Side 169 - THEOREM. 7?/6 convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.
Side 41 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Side 155 - AK. The two solids AG, AQ, having the same base AEHD are to each other as their altitudes AB, AO ; in like manner, the two solids AQ, AK, having the same base AOLE, are to each other as their altitudes AD, AM. Hence we have the two proportions, sol.
Side 86 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Side 282 - ... 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. ' Ans. 4.712364. 2. To find the length of an arc of 12° 10', or 12£°, the diameter being 20 feet.
Side 93 - ABC : FGH : : ACD : FHI. By the same mode of reasoning, we should find ACD : FHI : : ADE : FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents...